# Homework Help: Did I solve this equation right? (Bernoulli)

1. Feb 6, 2013

### 1MileCrash

1. The problem statement, all variables and given/known data

t^2y' + 2ty - y^3 = 0

2. Relevant equations

3. The attempt at a solution

y'+(2/t)y = (1/t^2)y^3

Let v = y^-2; then dv/dy = -2y^-3(dy/dt) and (dy/dt)=(-1/2)y^3(dv/dy)

Making that sub we have:

(-1/2)y^3(dv/dy)+(2/t)y = (1/t^2)y^3

(dv/dy)+(-4/t)y^-2 = (-2/t^2)
(dv/dy)+(-4/t)v = (-2/t^2)

My integrating factor is t^-4, giving me

v = (2/5)t^-1 + Ct^4
y^-2 = (2/5)t^-1 + Ct^4
y^2 = ((2/5)t^-1 + Ct^4)^-1

y= (+/-) (((2/5)t^-1 + Ct^4)^-1)^(1/2)

Wolfram gives something pretty irreconcilable.

2. Feb 7, 2013

### Karnage1993

Your solution is correct. Note that you can plug in both values for y back into the DE to see if the left hand side equals 0, which is does.

3. Feb 7, 2013

### 1MileCrash

Thanks for the response, much appreciated.

I'm having more trouble checking the validity of my solutions than finding them in the first place, turns into an algebraic mess.

For y' = ry-ky^2 (k r constant) I get:

y=((k/r)+ce^(-rt))^-1

And again wolfram gives something weird, and for

y' = ky - ny^3 I get

y= (+/-)sqrt((n/k)+ce^(-2kt)^-1)

Sorry again for the lack of latex, still doesn't work on my browser ( though admittedly if I knew the markup by memory without clicking the buttons it would work :/)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook